Given two integers N and X, where N is the number of nodes in an almost complete binary tree. The task is to find:
- The number of nodes that are at X distance from the root.
- The number of nodes that are at X distance from any leaf in its subtree.
Note: A Complete binary tree is a binary tree in which every level, except possibly the last, is completely filled and all nodes are as far left as possible.
Examples:
Input: N = 6, X = 0
Output:
1
3
Complete Binary Tree of 6 nodes is
1
/ \
2 3
/ \ /
4 5 6
Nodes that are at 0 distance from root = 1 (root itself).
Nodes that are at 0 distance from any of the leaf = 3 (all the leaves of the tree)
Input: N = 13, X = 1
Output:
2
4
Complete Binary Tree of 13 nodes.
1
/ \
2 3
/ \ / \
4 5 6 7
/ \ / \ / \
8 9 10 11 12 13
Nodes that are at 0 distance from root = 2 (node no. 2 and 3)
Nodes that are at 0 distance from any of the leaf = 4 (node no. 4, 5, 6 and 3)
Approach:
- Finding the number of nodes that are x distance away from the root is simple. We simply print the number of nodes at x-th height. We know that in a complete binary tree every level is complete, except possibly the last and all nodes are as far left as possible. So, the height h of a complete binary tree is calculated as floor(log2(n)) where n is the total number of nodes. Also, the number of nodes at i-th height will be 2i and the number of nodes that are at last level (i.e at height h) = 2h – (max_total_nodes – n), where 2h are maximum nodes at height h.
- Finding the number of nodes that are x distance away from any leaf in its subtree is a little bit tricky. It’s about observing the fact that if we have l leaf nodes then, ceil(l/2) nodes are 1 unit distance away from leaves, ceil(l/4) nodes are 2 units distance away from leaves …. till ceil(l/2i) nodes are i units distance away from leaves. We will first find the count of leaf nodes and apply the above method to find nodes that are x distance away from leaf.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the count// of the required nodesvoid countNodes(int N, int X){ // Height of the complete binary // tree with n nodes int height = floor(log2(N)); // If X > height then no node // can be present at that level if (X > height) { cout << "0\n0"; return; } // Corner case if (N == 1) { cout << "1\n1"; return; } // Maximum total nodes that are possible // in complete binary tree with height h int max_total_nodes = (1 << (height + 1)) - 1; // Nodes at the last level int nodes_last_level = (1 << height) - (max_total_nodes - N); // To store the count of nodes // x dist away from root int from_root; // To store the count of nodes // x dist away from leaf int from_leaf; // If X = h then print nodes at last level // else nodes at Xth level if (X == height) from_root = nodes_last_level; // 2^X else from_root = 1 << X; // Number of left leaf nodes at (h-1)th level // observe that if nodes are not present at last level // then there are a/2 leaf nodes at (h-1)th level int left_leaf_nodes = ((1 << height) - nodes_last_level) / 2; // If X = h then print leaf nodes at the last h level // + leaf nodes at (h-1)th level if (X == 0) { from_leaf = nodes_last_level + left_leaf_nodes; } else { // First calculate nodes for leaves present at // height h int i = X; while (nodes_last_level > 1 && i > 0) { nodes_last_level = ceil((float)nodes_last_level / (float)2); i--; } from_leaf = nodes_last_level; // Then calculate nodes for leaves present at height // h-1 i = X; while (left_leaf_nodes > 1 && i > 0) { left_leaf_nodes = ceil((float)left_leaf_nodes / (float)2); i--; } // Add both the results from_leaf += left_leaf_nodes; } cout << from_root << endl << from_leaf;}// Driver codeint main(){ int N = 38, X = 3; countNodes(N, X); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to find the count // of the required nodes static void countNodes(int N, int X) { // Height of the complete binary // tree with n nodes int height = (int)Math.floor(Math.log(N) / Math.log(2)); // If X > height then no node // can be present at that level if (X > height) { System.out.println("0\n0"); return; } // Corner case if (N == 1) { System.out.println("1\n1"); return; } // Maximum total nodes that are possible // in complete binary tree with height h int max_total_nodes = (1 << (height + 1)) - 1; // Nodes at the last level int nodes_last_level = (1 << height) - (max_total_nodes - N); // To store the count of nodes // x dist away from root int from_root; // To store the count of nodes // x dist away from leaf int from_leaf; // If X = h then print nodes at last level // else nodes at Xth level if (X == height) from_root = nodes_last_level; // 2^X else from_root = 1 << X; // Number of left leaf nodes at (h-1)th level // observe that if nodes are not present at last // level then there are a/2 leaf nodes at (h-1)th // level int left_leaf_nodes = ((1 << height) - nodes_last_level) / 2; // If X = h then print leaf nodes at the last h // level // + leaf nodes at (h-1)th level if (X == 0) { from_leaf = nodes_last_level + left_leaf_nodes; } else { // First calculate nodes for leaves present at // height h int i = X; while (nodes_last_level > 1 && i > 0) { nodes_last_level = (int)Math.ceil( (float)nodes_last_level / (float)2); i--; } from_leaf = nodes_last_level; // Then calculate nodes for leaves present at // height h-1 i = X; while (left_leaf_nodes > 1 && i > 0) { left_leaf_nodes = (int)Math.ceil( (float)left_leaf_nodes / (float)2); i--; } // Add both the results from_leaf += left_leaf_nodes; } System.out.println(from_root + "\n" + from_leaf); } // Driver Code public static void main(String[] args) { int N = 38, X = 3; countNodes(N, X); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 implementation of the approachfrom math import log2, ceil, floor# Function to find the count# of the required nodesdef countNodes(N, X): # Height of the complete binary # tree with n nodes height = floor(log2(N)) # If X > height then no node # can be present at that level if (X > height): print("0\n0") return # Corner case if (N == 1): print("1\n1") return # Maximum total nodes that are possible # in complete binary tree with height h max_total_nodes = (1 << (height + 1)) - 1 # Nodes at the last level nodes_last_level = (1 << height) - (max_total_nodes - N) # To store the count of nodes # x dist away from root from_root = 0 # To store the count of nodes # x dist away from leaf from_leaf = 0 # If X = h then print nodes at last level # else nodes at Xth level if (X == height): from_root = nodes_last_level # 2^X else: from_root = 1 << X # Number of left leaf nodes at (h-1)th level # observe that if nodes are not present at last level # then there are a/2 leaf nodes at (h-1)th level left_leaf_nodes = ((1 << height) - nodes_last_level) // 2 # If X = h then print leaf nodes at the last h level # + leaf nodes at (h-1)th level if (X == 0): from_leaf = nodes_last_level + left_leaf_nodes else: # First calculate nodes for leaves present at # height h i = X while (nodes_last_level > 1 and i > 0): nodes_last_level = ceil(nodes_last_level / 2) i-=1 from_leaf = nodes_last_level # Then calculate nodes for leaves present at height # h-1 i = X while (left_leaf_nodes > 1 and i > 0): left_leaf_nodes = ceil(left_leaf_nodes / 2) i -= 1 # Add both the results from_leaf += left_leaf_nodes print(from_root) print(from_leaf)# Driver codeif __name__ == '__main__': N, X = 38, 3 countNodes(N, X) # This code is contributed by mohit kumar 29. |
C#
// C# implementation of the approachusing System;class GFG{// Function to find the count// of the required nodesstatic void countNodes(int N, int X){ // Height of the complete binary // tree with n nodes int height = (int)Math.Floor(Math.Log(N) / Math.Log(2)); // If X > height then no node // can be present at that level if (X > height) { Console.Write("0\n0"); return; } // Corner case if (N == 1) { Console.Write("1\n1"); return; } // Maximum total nodes that are possible // in complete binary tree with height h int max_total_nodes = (1 << (height + 1)) - 1; // Nodes at the last level int nodes_last_level = (1 << height) - (max_total_nodes - N); // To store the count of nodes // x dist away from root int from_root; // To store the count of nodes // x dist away from leaf int from_leaf; // If X = h then print nodes at last level // else nodes at Xth level if (X == height) from_root = nodes_last_level; // 2^X else from_root = 1 << X; // Number of left leaf nodes at (h-1)th level // observe that if nodes are not present at last // level then there are a/2 leaf nodes at (h-1)th // level int left_leaf_nodes = ((1 << height) - nodes_last_level) / 2; // If X = h then print leaf nodes at the last h // level // + leaf nodes at (h-1)th level if (X == 0) { from_leaf = nodes_last_level + left_leaf_nodes; } else { // First calculate nodes for leaves present // at height h int i = X; while (nodes_last_level > 1 && i > 0) { nodes_last_level = (int)Math.Ceiling( (float)nodes_last_level / (float)2); i--; } from_leaf = nodes_last_level; // Then calculate nodes for leaves present at // height h-1 i = X; while (left_leaf_nodes > 1 && i > 0) { left_leaf_nodes = (int)Math.Ceiling( (float)left_leaf_nodes / (float)2); i--; } // Add both the results from_leaf += left_leaf_nodes; } Console.Write(from_root + "\n" + from_leaf);}// Driver Codepublic static void Main(){ int N = 38, X = 3; countNodes(N, X);}}// This code is contributed by subham348 |
Javascript
<script>// Javascript implementation of the approach// Function to find the count// of the required nodesfunction countNodes(N, X){ // Height of the complete binary // tree with n nodes let height = Math.floor(Math.log(N) / Math.log(2)); // If X > height then no node // can be present at that level if (X > height) { document.write("0<br>0"); return; } // Corner case if (N == 1) { document.write("1<br>1"); return; } // Maximum total nodes that are possible // in complete binary tree with height h let max_total_nodes = (1 << (height + 1)) - 1; // Nodes at the last level let nodes_last_level = (1 << height) - (max_total_nodes - N); // To store the count of nodes // x dist away from root let from_root; // To store the count of nodes // x dist away from leaf let from_leaf; // If X = h then print nodes at last level // else nodes at Xth level if (X == height) from_root = nodes_last_level; // 2^X else from_root = 1 << X; // Number of left leaf nodes at (h-1)th level // observe that if nodes are not present at last // level then there are a/2 leaf nodes at (h-1)th // level let left_leaf_nodes = Math.floor( ((1 << height) - nodes_last_level) / 2); // If X = h then print leaf nodes at the last h // level // + leaf nodes at (h-1)th level if (X == 0) { from_leaf = nodes_last_level + left_leaf_nodes; } else { // First calculate nodes for leaves // present at height h let i = X; while (nodes_last_level > 1 && i > 0) { nodes_last_level = Math.floor(Math.ceil( nodes_last_level / 2)); i--; } from_leaf = nodes_last_level; // Then calculate nodes for leaves present at // height h-1 i = X; while (left_leaf_nodes > 1 && i > 0) { left_leaf_nodes = Math.floor(Math.ceil( left_leaf_nodes / 2)); i--; } // Add both the results from_leaf += left_leaf_nodes; } document.write(from_root + "<br>" + from_leaf);}// Driver Codelet N = 38, X = 3;countNodes(N, X);// This code is contributed by unknown2108</script> |
Output:
8 3
Time Complexity: O(log (N) )
Auxiliary Space: O(1), since no extra space has been taken.
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